NumPy矩阵与数组类的乘法有何不同?

问题:NumPy矩阵与数组类的乘法有何不同?

numpy文档建议使用数组而不是矩阵来处理矩阵。但是,与八度(我直到最近才使用)不同,*不执行矩阵乘法,您需要使用函数matrixmultipy()。我觉得这使代码非常不可读。

是否有人分享我的观点并找到了解决方案?

The numpy docs recommend using array instead of matrix for working with matrices. However, unlike octave (which I was using till recently), * doesn’t perform matrix multiplication, you need to use the function matrixmultipy(). I feel this makes the code very unreadable.

Does anybody share my views, and has found a solution?


回答 0

避免使用的主要原因 matrix该类的是:a)本质上是二维的,并且b)与“常规” numpy数组相比,存在额外的开销。如果您要做的只是线性代数,那么请务必使用矩阵类…就我个人而言,我发现它比它值得的麻烦更多。

对于数组(Python 3.5之前的版本),请使用dot代替matrixmultiply

例如

import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)

或在新版本的numpy中,只需使用 x.dot(y)

就个人而言,我发现它比*表示矩阵乘法的运算符更具可读性…

对于Python 3.5中的数组,请使用x @ y

The main reason to avoid using the matrix class is that a) it’s inherently 2-dimensional, and b) there’s additional overhead compared to a “normal” numpy array. If all you’re doing is linear algebra, then by all means, feel free to use the matrix class… Personally I find it more trouble than it’s worth, though.

For arrays (prior to Python 3.5), use dot instead of matrixmultiply.

E.g.

import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)

Or in newer versions of numpy, simply use x.dot(y)

Personally, I find it much more readable than the * operator implying matrix multiplication…

For arrays in Python 3.5, use x @ y.


回答 1

与在NumPy 矩阵上进行操作相比,在NumPy 数组上进行操作要了解的关键事项是:

  • NumPy矩阵是NumPy数组的子类

  • NumPy 数组操作是基于元素的(一旦考虑了广播)

  • NumPy 矩阵运算遵循线性代数的一般规则

一些代码片段来说明:

>>> from numpy import linalg as LA
>>> import numpy as NP

>>> a1 = NP.matrix("4 3 5; 6 7 8; 1 3 13; 7 21 9")
>>> a1
matrix([[ 4,  3,  5],
        [ 6,  7,  8],
        [ 1,  3, 13],
        [ 7, 21,  9]])

>>> a2 = NP.matrix("7 8 15; 5 3 11; 7 4 9; 6 15 4")
>>> a2
matrix([[ 7,  8, 15],
        [ 5,  3, 11],
        [ 7,  4,  9],
        [ 6, 15,  4]])

>>> a1.shape
(4, 3)

>>> a2.shape
(4, 3)

>>> a2t = a2.T
>>> a2t.shape
(3, 4)

>>> a1 * a2t         # same as NP.dot(a1, a2t) 
matrix([[127,  84,  85,  89],
        [218, 139, 142, 173],
        [226, 157, 136, 103],
        [352, 197, 214, 393]])

但是如果将以下两个NumPy矩阵转换为数组,则此操作将失败:

>>> a1 = NP.array(a1)
>>> a2t = NP.array(a2t)

>>> a1 * a2t
Traceback (most recent call last):
   File "<pyshell#277>", line 1, in <module>
   a1 * a2t
   ValueError: operands could not be broadcast together with shapes (4,3) (3,4) 

尽管使用NP.dot语法可以处理数组 ; 该操作类似于矩阵乘法:

>> NP.dot(a1, a2t)
array([[127,  84,  85,  89],
       [218, 139, 142, 173],
       [226, 157, 136, 103],
       [352, 197, 214, 393]])

那么您是否需要NumPy矩阵?即,NumPy数组是否足以进行线性代数计算(前提是您知道正确的语法,即NP.dot)?

规则似乎是,如果参数(数组)的形状(mxn)与给定的线性代数运算兼容,那么您就可以了,否则,NumPy抛出。

我遇到的唯一exceptions(可能还有其他exceptions)是计算矩阵逆

下面是我称为纯线性代数运算(实际上是从Numpy的线性代数模块)并传递给NumPy数组的代码片段

数组的行列式

>>> m = NP.random.randint(0, 10, 16).reshape(4, 4)
>>> m
array([[6, 2, 5, 2],
       [8, 5, 1, 6],
       [5, 9, 7, 5],
       [0, 5, 6, 7]])

>>> type(m)
<type 'numpy.ndarray'>

>>> md = LA.det(m)
>>> md
1772.9999999999995

特征向量/特征值对:

>>> LA.eig(m)
(array([ 19.703+0.j   ,   0.097+4.198j,   0.097-4.198j,   5.103+0.j   ]), 
array([[-0.374+0.j   , -0.091+0.278j, -0.091-0.278j, -0.574+0.j   ],
       [-0.446+0.j   ,  0.671+0.j   ,  0.671+0.j   , -0.084+0.j   ],
       [-0.654+0.j   , -0.239-0.476j, -0.239+0.476j, -0.181+0.j   ],
       [-0.484+0.j   , -0.387+0.178j, -0.387-0.178j,  0.794+0.j   ]]))

矩阵范数

>>>> LA.norm(m)
22.0227

qr因式分解

>>> LA.qr(a1)
(array([[ 0.5,  0.5,  0.5],
        [ 0.5,  0.5, -0.5],
        [ 0.5, -0.5,  0.5],
        [ 0.5, -0.5, -0.5]]), 
 array([[ 6.,  6.,  6.],
        [ 0.,  0.,  0.],
        [ 0.,  0.,  0.]]))

矩阵等级

>>> m = NP.random.rand(40).reshape(8, 5)
>>> m
array([[ 0.545,  0.459,  0.601,  0.34 ,  0.778],
       [ 0.799,  0.047,  0.699,  0.907,  0.381],
       [ 0.004,  0.136,  0.819,  0.647,  0.892],
       [ 0.062,  0.389,  0.183,  0.289,  0.809],
       [ 0.539,  0.213,  0.805,  0.61 ,  0.677],
       [ 0.269,  0.071,  0.377,  0.25 ,  0.692],
       [ 0.274,  0.206,  0.655,  0.062,  0.229],
       [ 0.397,  0.115,  0.083,  0.19 ,  0.701]])
>>> LA.matrix_rank(m)
5

矩阵条件

>>> a1 = NP.random.randint(1, 10, 12).reshape(4, 3)
>>> LA.cond(a1)
5.7093446189400954

反演需要一个NumPy矩阵

>>> a1 = NP.matrix(a1)
>>> type(a1)
<class 'numpy.matrixlib.defmatrix.matrix'>

>>> a1.I
matrix([[ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028]])
>>> a1 = NP.array(a1)
>>> a1.I

Traceback (most recent call last):
   File "<pyshell#230>", line 1, in <module>
   a1.I
   AttributeError: 'numpy.ndarray' object has no attribute 'I'

但是Moore-Penrose伪逆似乎工作得很好

>>> LA.pinv(m)
matrix([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
        [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
        [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
        [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
        [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

>>> m = NP.array(m)

>>> LA.pinv(m)
array([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
       [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
       [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
       [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
       [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

the key things to know for operations on NumPy arrays versus operations on NumPy matrices are:

  • NumPy matrix is a subclass of NumPy array

  • NumPy array operations are element-wise (once broadcasting is accounted for)

  • NumPy matrix operations follow the ordinary rules of linear algebra

some code snippets to illustrate:

>>> from numpy import linalg as LA
>>> import numpy as NP

>>> a1 = NP.matrix("4 3 5; 6 7 8; 1 3 13; 7 21 9")
>>> a1
matrix([[ 4,  3,  5],
        [ 6,  7,  8],
        [ 1,  3, 13],
        [ 7, 21,  9]])

>>> a2 = NP.matrix("7 8 15; 5 3 11; 7 4 9; 6 15 4")
>>> a2
matrix([[ 7,  8, 15],
        [ 5,  3, 11],
        [ 7,  4,  9],
        [ 6, 15,  4]])

>>> a1.shape
(4, 3)

>>> a2.shape
(4, 3)

>>> a2t = a2.T
>>> a2t.shape
(3, 4)

>>> a1 * a2t         # same as NP.dot(a1, a2t) 
matrix([[127,  84,  85,  89],
        [218, 139, 142, 173],
        [226, 157, 136, 103],
        [352, 197, 214, 393]])

but this operations fails if these two NumPy matrices are converted to arrays:

>>> a1 = NP.array(a1)
>>> a2t = NP.array(a2t)

>>> a1 * a2t
Traceback (most recent call last):
   File "<pyshell#277>", line 1, in <module>
   a1 * a2t
   ValueError: operands could not be broadcast together with shapes (4,3) (3,4) 

though using the NP.dot syntax works with arrays; this operations works like matrix multiplication:

>> NP.dot(a1, a2t)
array([[127,  84,  85,  89],
       [218, 139, 142, 173],
       [226, 157, 136, 103],
       [352, 197, 214, 393]])

so do you ever need a NumPy matrix? ie, will a NumPy array suffice for linear algebra computation (provided you know the correct syntax, ie, NP.dot)?

the rule seems to be that if the arguments (arrays) have shapes (m x n) compatible with the a given linear algebra operation, then you are ok, otherwise, NumPy throws.

the only exception i have come across (there are likely others) is calculating matrix inverse.

below are snippets in which i have called a pure linear algebra operation (in fact, from Numpy’s Linear Algebra module) and passed in a NumPy array

determinant of an array:

>>> m = NP.random.randint(0, 10, 16).reshape(4, 4)
>>> m
array([[6, 2, 5, 2],
       [8, 5, 1, 6],
       [5, 9, 7, 5],
       [0, 5, 6, 7]])

>>> type(m)
<type 'numpy.ndarray'>

>>> md = LA.det(m)
>>> md
1772.9999999999995

eigenvectors/eigenvalue pairs:

>>> LA.eig(m)
(array([ 19.703+0.j   ,   0.097+4.198j,   0.097-4.198j,   5.103+0.j   ]), 
array([[-0.374+0.j   , -0.091+0.278j, -0.091-0.278j, -0.574+0.j   ],
       [-0.446+0.j   ,  0.671+0.j   ,  0.671+0.j   , -0.084+0.j   ],
       [-0.654+0.j   , -0.239-0.476j, -0.239+0.476j, -0.181+0.j   ],
       [-0.484+0.j   , -0.387+0.178j, -0.387-0.178j,  0.794+0.j   ]]))

matrix norm:

>>>> LA.norm(m)
22.0227

qr factorization:

>>> LA.qr(a1)
(array([[ 0.5,  0.5,  0.5],
        [ 0.5,  0.5, -0.5],
        [ 0.5, -0.5,  0.5],
        [ 0.5, -0.5, -0.5]]), 
 array([[ 6.,  6.,  6.],
        [ 0.,  0.,  0.],
        [ 0.,  0.,  0.]]))

matrix rank:

>>> m = NP.random.rand(40).reshape(8, 5)
>>> m
array([[ 0.545,  0.459,  0.601,  0.34 ,  0.778],
       [ 0.799,  0.047,  0.699,  0.907,  0.381],
       [ 0.004,  0.136,  0.819,  0.647,  0.892],
       [ 0.062,  0.389,  0.183,  0.289,  0.809],
       [ 0.539,  0.213,  0.805,  0.61 ,  0.677],
       [ 0.269,  0.071,  0.377,  0.25 ,  0.692],
       [ 0.274,  0.206,  0.655,  0.062,  0.229],
       [ 0.397,  0.115,  0.083,  0.19 ,  0.701]])
>>> LA.matrix_rank(m)
5

matrix condition:

>>> a1 = NP.random.randint(1, 10, 12).reshape(4, 3)
>>> LA.cond(a1)
5.7093446189400954

inversion requires a NumPy matrix though:

>>> a1 = NP.matrix(a1)
>>> type(a1)
<class 'numpy.matrixlib.defmatrix.matrix'>

>>> a1.I
matrix([[ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028]])
>>> a1 = NP.array(a1)
>>> a1.I

Traceback (most recent call last):
   File "<pyshell#230>", line 1, in <module>
   a1.I
   AttributeError: 'numpy.ndarray' object has no attribute 'I'

but the Moore-Penrose pseudoinverse seems to works just fine

>>> LA.pinv(m)
matrix([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
        [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
        [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
        [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
        [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

>>> m = NP.array(m)

>>> LA.pinv(m)
array([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
       [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
       [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
       [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
       [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

回答 2

在3.5中,Python终于有了一个矩阵乘法运算符。语法为a @ b

In 3.5, Python finally got a matrix multiplication operator. The syntax is a @ b.


回答 3

在处理数组和处理矩阵时,点运算符会给出不同的答案。例如,假设以下内容:

>>> a=numpy.array([1, 2, 3])
>>> b=numpy.array([1, 2, 3])

让我们将它们转换成矩阵:

>>> am=numpy.mat(a)
>>> bm=numpy.mat(b)

现在,我们可以看到两种情况的不同输出:

>>> print numpy.dot(a.T, b)
14
>>> print am.T*bm
[[1.  2.  3.]
 [2.  4.  6.]
 [3.  6.  9.]]

There is a situation where the dot operator will give different answers when dealing with arrays as with dealing with matrices. For example, suppose the following:

>>> a=numpy.array([1, 2, 3])
>>> b=numpy.array([1, 2, 3])

Lets convert them into matrices:

>>> am=numpy.mat(a)
>>> bm=numpy.mat(b)

Now, we can see a different output for the two cases:

>>> print numpy.dot(a.T, b)
14
>>> print am.T*bm
[[1.  2.  3.]
 [2.  4.  6.]
 [3.  6.  9.]]

回答 4

来自http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html的参考

…,使用的numpy.matrix气馁,因为它增加了什么,无法与2D来完成numpy.ndarray对象,并可能导致混乱,其中正在使用的类。例如,

>>> import numpy as np
>>> from scipy import linalg
>>> A = np.array([[1,2],[3,4]])
>>> A
    array([[1, 2],
           [3, 4]])
>>> linalg.inv(A)
array([[-2. ,  1. ],
      [ 1.5, -0.5]])
>>> b = np.array([[5,6]]) #2D array
>>> b
array([[5, 6]])
>>> b.T
array([[5],
      [6]])
>>> A*b #not matrix multiplication!
array([[ 5, 12],
      [15, 24]])
>>> A.dot(b.T) #matrix multiplication
array([[17],
      [39]])
>>> b = np.array([5,6]) #1D array
>>> b
array([5, 6])
>>> b.T  #not matrix transpose!
array([5, 6])
>>> A.dot(b)  #does not matter for multiplication
array([17, 39])

scipy.linalg操作可以同等地应用于numpy.matrix或2D numpy.ndarray对象。

Reference from http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html

…, the use of the numpy.matrix class is discouraged, since it adds nothing that cannot be accomplished with 2D numpy.ndarray objects, and may lead to a confusion of which class is being used. For example,

>>> import numpy as np
>>> from scipy import linalg
>>> A = np.array([[1,2],[3,4]])
>>> A
    array([[1, 2],
           [3, 4]])
>>> linalg.inv(A)
array([[-2. ,  1. ],
      [ 1.5, -0.5]])
>>> b = np.array([[5,6]]) #2D array
>>> b
array([[5, 6]])
>>> b.T
array([[5],
      [6]])
>>> A*b #not matrix multiplication!
array([[ 5, 12],
      [15, 24]])
>>> A.dot(b.T) #matrix multiplication
array([[17],
      [39]])
>>> b = np.array([5,6]) #1D array
>>> b
array([5, 6])
>>> b.T  #not matrix transpose!
array([5, 6])
>>> A.dot(b)  #does not matter for multiplication
array([17, 39])

scipy.linalg operations can be applied equally to numpy.matrix or to 2D numpy.ndarray objects.


回答 5

这个技巧可能就是您想要的。这是一种简单的运算符重载。

然后,您可以使用类似建议的Infix类的东西:

a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b

This trick could be what you are looking for. It is a kind of simple operator overload.

You can then use something like the suggested Infix class like this:

a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b

回答 6

来自PEP 465的相关报价 @ petr-viktorin提到的用于矩阵乘法的专用中缀运算符,阐明了OP遇到的问题:

numpy提供了两种使用不同__mul__方法的不同类型。对于numpy.ndarray对象,*执行元素乘法,矩阵乘法必须使用函数调用(numpy.dot)。对于numpy.matrix对象,*执行矩阵乘法,而元素乘法则需要函数语法。使用编写代码numpy.ndarray效果很好。使用编写代码numpy.matrix也可以。但是,一旦我们尝试将这两段代码集成在一起,麻烦就会开始。预期为ndarray并得到matrix或相反的代码可能会崩溃或返回错误的结果

@infix运算符的引入应有助于统一和简化python矩阵代码。

A pertinent quote from PEP 465 – A dedicated infix operator for matrix multiplication , as mentioned by @petr-viktorin, clarifies the problem the OP was getting at:

[…] numpy provides two different types with different __mul__ methods. For numpy.ndarray objects, * performs elementwise multiplication, and matrix multiplication must use a function call (numpy.dot). For numpy.matrix objects, * performs matrix multiplication, and elementwise multiplication requires function syntax. Writing code using numpy.ndarray works fine. Writing code using numpy.matrix also works fine. But trouble begins as soon as we try to integrate these two pieces of code together. Code that expects an ndarray and gets a matrix, or vice-versa, may crash or return incorrect results

The introduction of the @ infix operator should help to unify and simplify python matrix code.


回答 7

函数matmul(自numpy 1.10.1起)对两种类型均适用,并以numpy矩阵类返回结果:

import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))

输出:

(matrix([[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9],
        [10, 11, 12]]), <class 'numpy.matrixlib.defmatrix.matrix'>)
(array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]]), <type 'numpy.ndarray'>)
(matrix([[ 6,  6,  6,  6],
        [15, 15, 15, 15],
        [24, 24, 24, 24],
        [33, 33, 33, 33]]), <class 'numpy.matrixlib.defmatrix.matrix'>)

由于python 3.5 如前所述,您还可以使用新的矩阵乘法运算符,@例如

C = A @ B

并获得与上述相同的结果。

Function matmul (since numpy 1.10.1) works fine for both types and return result as a numpy matrix class:

import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))

Output:

(matrix([[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9],
        [10, 11, 12]]), <class 'numpy.matrixlib.defmatrix.matrix'>)
(array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]]), <type 'numpy.ndarray'>)
(matrix([[ 6,  6,  6,  6],
        [15, 15, 15, 15],
        [24, 24, 24, 24],
        [33, 33, 33, 33]]), <class 'numpy.matrixlib.defmatrix.matrix'>)

Since python 3.5 as mentioned early you also can use a new matrix multiplication operator @ like

C = A @ B

and get the same result as above.